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3 years ago

Which line from the prologue of Romeo and Juliet reveals the ending of the play?

Physics

2 answers:

KonstantinChe [14]3 years ago

7 0

Answer:

" A pair of star crossed lovers who take their life,

Whose misadventured piteous overthrows

Doth with their death bury their parents' strife"

Explanation:

The prologue is essential for the play for the following reasons:

-Sets the play in Verona

-Exposes Montagues and Capulates' rivalry

-Romeo and Juliet (the pair of star crossed lovers who take their life), come from these families.

ira [324]3 years ago

6 0

A pair of star-cross'd lovers take their life

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The maximum speed limit on interstate 10 is 75 miles per hour. how many meters per second is this

Dvinal [7]

Answer:

Explanation:

Given the maximum speed limit on interstate 10 as 75 miles per hour, to get the speed in meter per seconds, we need to convert the given speed to meter per seconds.

Using the conversion 1 mile = 1609.34m and 1 hour = 3600 seconds

75 miles perhour = 75miles/1 hour

75miles/1 hour (in m/s) = 75miles*1609.34m* 1 hour/1mile * 1 hour * 3600s *

= 75 *1609.34m* 1 /1 * 1 * 3600s

= 120,700.5m/3600s

= 33.53m/s

Hence the maximum speed limit on interstate 10 in metre per seconds is 33.53m/s

8 0

3 years ago

A solar eclipse that occurs when the new moon is too far from earth to completely cover the sun can be either a partial solar ec

Sonja [21]

1) A solar eclipse that occurs when the new moon is too far from earth to completely cover the sun can be either a partial solar eclipse or an -->
Answer: ANULAR ECLIPSE. Since the moon is too far, it will cover only a part of the sun, and only the external ring of the moon will be visible; this is called anular eclipse.

2) anyone looking from the night side of earth can, in principle, see a -->
Answer: LUNAR ECLIPSE. If the moon is the right position, and the Earth's shadow covers partially or totally the moon, then a lunar eclipse occurs.

3) during some lunar eclipses, the moon's appearance changes only slightly, because it passes only through the part of earth's shadow called the -->
Answer: PENUMBRA.

4) a ... can occur only when the moon is new and has an angular size larger than the sun in the sky -->
Answer: TOTAL SOLAR ECLIPSE. When the moon is new, it means it is between the sun and the Earth, and its dark side faces the Earth. If the moon's angular size is also larger than the sun angular size, than it will completely cover the sun, and a total solar eclipse occurs.

5) a partial lunar eclipse begins when the moon first touches earth's -->
Answer: SHADOW. The Earth's shadow will start to cover the moon, and partial lunar eclipse will start.

6) a point at which the moon crosses earth's orbital plane is called a(n) -->
Answer: NODE. Eclipses occur only when the Moon is at or close to a node, otherwise sun, earth and moon are not "aligned".

5 0

3 years ago

Calculate the work done if a boy lifts a bag of cement 500N to the floor of a lorry 2.5m above the ground

eduard

Answer:

W = 1250 J = 1.25 KJ

Explanation:

The work done by the boy is due to the change in the position of the cement vertically. Hence, the work done in this case will be equal to the potential energy of the cement:

$W = P.E\\W = mgh$

where,

W = Work done = ?

mg = W = weight of cement = 500 N

h = height covered = 2.5 m

Therefore,

$W = (500\ N)(2.5\ m)$

W = 1250 J = 1.25 KJ

4 0

2 years ago

A horizontal pipe of diameter 1.03m has a smooth constriction to a section of diameter 0.618 m. The density of oil flowing in th

vodka [1.7K]

Velocity of the oil in the pipe: 0.76 m/s, in the constricted section: 5.87 m/s

Explanation:

We can solve this problem by using Bernoulli's equation:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$ (1)

where

$p_1 = 7340 N/m^2$ is the pressure in the pipe

$p_2 = 5505 N/m^2$ is the pressure in the constricted section

$\rho = 821 kg/m^3$ is the density of the oil

$v_1$ is the velocity of the oil in the pipe

$v_2$ is the velocity of the oil in the constricted section

Also, according to the continuity equation,

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the pipe, with

$r_1 = \frac{1.03}{2}=0.515 m$ is the radius

$A_2 = \pi r_2^2$ is the cross-sectional area of the constricted section, with

$r_2=\frac{0.618}{2}=0.309 m$ is the radius

So the equation becomes

$r_1^2 v_1 = r_2^2 v_2$

So we can write

$v_2=\frac{r_1^2}{r_2^2}v_1$

Substituting into eq.(1),

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2$

And solving the equation for $v_1$ :

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho \frac{r_1^4}{r_2^4}v_1^2\\v_1=\sqrt{\frac{p_2-p_1}{\frac{1}{2}\rho-\frac{1}{2}\rho \frac{r_1^4}{r_2^4}}}=0.76 m/s$